3

Recall that the closed interval $[a,b]$={$x \in \mathbb{R} \vert a\le x \le b$}

prove that $\vert [2,5] \vert = \vert [-2,3] \vert$

(cardinality of [a,b] not absolute value)

proof: define $f:[2,5] \rightarrow [-2,3]$ by $f(x)=c+ \frac{d-c}{b-a} (x-a)=-2+ \frac{5}{3}(x-2)$

show that $f$ is injective:

let $a,b \in [2,5]$ such that $f(a)=f(b)$

then

$-2+\frac{5}{3}(a-2)=-2+\frac{5}{3}(b-2)$

$\frac{5a-10}{3}=\frac{5b-10}{3}$

$\frac{5a}{3}=\frac{5b}{3}$ $\rightarrow$ $a=b$

so f is injective.

now when i try to show that f is surjective, i let $y\in [-2,3]$

then if $x\in [2,5]$ such that $x=\frac{3(y+2)+10)}{15}$

we see that $f(x)=-2+\frac{5}{3}((\frac{3(y+2)+10}{15})-2)$

basically trying to show that this simplifies to $f(x)=y$ and hence is surjective, was gettng way too tedious and wasnt looking very promising. Can someone please show me where I went wrong here?

kr1s
  • 311
  • What is $|[2,5]|=|[-2,3]|$ supposed to mean? How do you define the "absolute value" of an intervall $[a,b]$ – Cornman Oct 01 '17 at 00:44
  • 1
    $\vert[a,b] \vert$ is cardinality of [a,b] – kr1s Oct 01 '17 at 00:49
  • Both sets are infinite... – Juniven Acapulco Oct 01 '17 at 00:52
  • yes but i need to show that theres a one-one correspondence between every value of the domain and codomain (bijection) or maybe im hitting a brick wall because im trying to prove this to be true when its actually false, but if both sets are infinite then shouldnt they have equal cardinality? – kr1s Oct 01 '17 at 01:01
  • Your bijection is affine (of the form ax+b) so it is trivially shown to be invertible. –  Oct 01 '17 at 01:05

1 Answers1

1

You have $$f(x) = \frac{5}{3}(x - 2) - 2$$ Let $y = f(x)$, then solve for $x$. \begin{align*} y & = \frac{5}{3}(x - 2) - 2\\ y + 2 & = \frac{5}{3}(x - 2)\\ \frac{3}{5}(y + 2) & = x - 2\\ \frac{3}{5}(y + 2) + 2 & = x \end{align*} Substituting this value for $x$ in the function $f$ yields \begin{align*} f\left(\frac{3}{5}(y + 2) + 2\right) & = \frac{5}{3}\left[\frac{3}{5}(y + 2) + 2 - 2\right] - 2\\ & = \frac{5}{3}\left[\frac{3}{5}(y + 2)\right] - 2\\ & = y + 2 - 2\\ & = y \end{align*} so $f$ is surjective.

Since you did not show your work, I am not sure how you obtained a $15$ in your denominator. However, that was your error.

Note: Another approach to showing that a bijection exists would be to solve for $$g(x) = \frac{3}{5}(x + 2) + 2$$ then showing that $g = f^{-1}$ by demonstrating that \begin{align*} (g \circ f)(x) & = x && \text{for each $x \in \text{Dom}_f = [2, 5]$}\\ (f \circ g)(x) & = x && \text{for each $x \in \text{Dom}_g = [-2, 3]$} \end{align*}

N. F. Taussig
  • 76,571
  • 1
    it looks as if i made a simple mistake with my algebra, thankyou! also thankyou for the alternative method, it seems much simpler. – kr1s Oct 02 '17 at 01:18